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![\frac{(2x+1)(3x-2)}{24x^{2}-4x-8} = \frac{(2x+1)(3x+2)}{4(6x^{2}-x-2)} = \frac{(2x+1)(3x+2)}{4(6x^{2}-4x+3x-2)} = \frac{(2x+1)(3x+2)}{4[2x(3x-2)+(3x-2)]} = \\ \\ = \frac{(2x+1)(3x+2)}{4(2x+1)(3x-2)} = \frac{3x+2}{4(3x-2)} = \frac{3x+2}{12x-8}](/tpl/images/0245/7685/b34e1.png)
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